Matrix expression for elements of $SO(3)$ Hi all. Is there any explicit matrix expression for a general element of the special orthogonal group $SO(3)$? I have been searching texts and net both, but could not find it. Kindly provide any references.
 A: Here is the standard quaternion answer: Given $(a,b,c,d)$ such that $a^2+b^2+c^2+d^2=1$, the matrix
$$\begin{pmatrix}
a^2+b^2-c^2-d^2&2bc-2ad        &2bd+2ac        \\
2bc+2ad        &a^2-b^2+c^2-d^2&2cd-2ab        \\
2bd-2ac        &2cd+2ab        &a^2-b^2-c^2+d^2\\
\end{pmatrix}$$
is a rotation and every rotation matrix is of this form. Note that $(a,b,c,d)$ and $(-a, -b,-c,-d)$ give the same rotation.
A: There is a good way to derive the sort of thing you're looking for: use the double cover $SU(2) \to SO(3)$.  $SU(2)$ is diffeomorphic to the 3-sphere $S^3 \subseteq \mathbb{C}^2$
$$ SU(2) = \left\{ \begin{pmatrix} a & -\overline{b} \\ b & \overline{a} \end{pmatrix} : |a|^2 + |b|^2 = 1 \right\} $$
Now $SU(2)$ acts on its Lie algebra $\mathfrak{su}_2$ (which is 3-dimensional) by conjugation.  This action preserves the inner product
$$ \langle X, Y \rangle = - \frac12 \mathrm{tr}(XY) = \frac12 \mathrm{tr}(X^*Y)$$
(which is a scalar multiple of the Killing form of $\mathfrak{su}_2$, FYI) and hence this gives a homomorphism
$SU(2) \to SO(\mathfrak{su}_2) \simeq SO(3)$.  (A priori this gives a map to $O(3)$, but $SU(2)$ is connected so the image lands in $SO(3)$.
Now consider the orthonormal basis for $\mathfrak{su}_2$ given by
$$
e_1 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, e_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, e_3= \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix},
$$
let 
$$
x = \begin{pmatrix} a & -\overline{b} \\ b & \overline{a}\end{pmatrix},$$
and write down the adjoint action of $x$ on $e_1,e_2,e_3$.  For instance you get
$$ 
\begin{align}
xe_1x^{-1} & = \begin{pmatrix} i(|a|^2 - |b|^2) & 2ia\overline{b} \\ 2i\overline{a} b & -i(|a|^2 - |b|^2) \end{pmatrix} \\
& = (|a|^2 - |b|^2)e_1 + i(a\overline{b} - \overline{a}b)e_2 + (a \overline{b} + \overline{a}b)e_3.
\end{align}$$
This gives you the first column of the matrix representation conjugation by $x$.  I'll leave the others to you.  But this way you can see where the formulas come from.
This gives exactly David Speyer's answer (possibly modulo some re-ordering of the basis).  His four real numbers $a,b,c,d$ would correspond to my complex numbers $a,b$ via 
$$a_{mine} = a + i b, \quad b_{mine} = c + id.$$
A: To give something explicit in sine and cosine,
$$   
 \left(  \begin{array}{ccc}
\cos\theta \cos\psi & -\cos\phi \sin\psi + \sin\phi \sin\theta \cos\psi &   \sin\phi \sin\psi + \cos\phi \sin\theta \cos\psi \\\
\cos\theta \sin\psi &  \cos\phi \cos\psi + \sin\phi \sin\theta \sin\psi & -\sin\phi \cos\psi + \cos\phi \sin\theta \sin\psi \\\
-\sin\theta             &  \sin\phi \cos\theta                                          &   \cos\phi \cos\theta 
\end{array} 
  \right)  
  $$
Note that three parameters are required. In odd dimension, there is a real eigenvalue. For $SO_n$ this eigenvalue is $+1.$ So there is a fixed vector in some direction. It takes two parameters to specify this point on the unit sphere. The Lie group element is then a rotation around this point. So it takes a third parameter specifying the amount of rotation about that axis.  
